Question: Solve for $r$. Give an exact answer. $\dfrac12r-3=3\left(4-\dfrac32r\right)$ $r =$
Explanation: We need to manipulate the equation to get $ r $ by itself. $\begin{aligned} \dfrac12r-3 &= 3\left(4-\dfrac32r\right) \\\\ \dfrac12r-3 &= 12-\dfrac92r~~~~~~~~~~\gray{\text{Distribute}}\\\\ \dfrac12r-3{+\dfrac92r} &= 12-\dfrac92r{+\dfrac92r} ~~~~~~~~~~\gray{\text{Add }\dfrac92r \text{ to each side}}\\\\ 5r-3 &=12 ~~~~~~~~~~\gray{\text{Combine like terms}}\\\\ 5r-3{+3} &=12 {+3} ~~~~~~~~~~\gray{\text{Add 3 to each side}}\\\\ 5r &=15 ~~~~~~~~~~\gray{\text{Combine like terms}}\\\\ \dfrac{5r}{{5}}&= \dfrac{15}{{5}}~~~~~~~~~~\gray{\text{Divide each side by 5}} \\\\ r &= {3} ~~~~~~~~~~\gray{\text{Simplify}}\\\\ \end{aligned}$ The answer: $ r = { 3 }~~~~~~~~$ [Let's check our work!]